VIẾT CÁC BIỂU THỨC SAU DƯỚI DẠNG BÌNH PHƯƠNG CỦA 1 TỔNG (HOẶC 1 HIỆU)
x2 + 10x + 25
x2 - 6x + 9
x2 - x + 1/4
x2y4 - 2xy2 + 1
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a)x2-6x+9
=x2-2.x.3+32
=(x-3)2
b)4x2+4x+1
=(2x)2+2.2x.1+12
=(2x+1)2
c)4x2+12xy+9y2
=(2x)2+2.2x.3y+(3y)2
=(2x+3y)2
d)4x4-4x2+4
=(2x2)2-2.2x2.2+22
=(2x2-2)2
Bài 1:
a) \(a^2-6a+9=\left(a-3\right)^2\)
b) \(\dfrac{1}{4}x^2+2xy^2+4y^4=\left(\dfrac{1}{2}x+2y^2\right)^2\)
Bài 2:
a) \(\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\)
\(\Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\)
b) \(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)
a) ( x + 1 ) 2 . b) ( x – 4 ) 2 .
c) x 2 4 + x + 1 ; d) ( 2 x – 2 y ) 2 .
a) ( 2 x + 1 ) 2 . b) ( 3 x – 2 ) 2 .
c) 1 2 ab 2 + 1 2 . d) ( 4 uv 2 – 1 ) 2 .
a) \(x^2+4x+4\)
\(=x^2+2\cdot2\cdot x+2^2\)
\(=\left(x+2\right)^2\)
b) \(4x^2-4x+1\)
\(=\left(2x\right)^2-2\cdot2x\cdot1+1^2\)
\(=\left(2x-1\right)^2\)
c) \(x^2-x+\dfrac{1}{4}\)
\(=x^2-2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}\right)^2\)
d) \(4\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=\left[2\left(x+y\right)\right]^2-2\cdot2\left(x+y\right)\cdot1+1^2\)
\(=\left[2\left(x+y\right)-1\right]^2\)
\(=\left(2x+2y-1\right)^2\)
\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)
a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)
này mình có vài câu không làm được, xin lỗi bạn nha
\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)
\(x^2+10x+25\)
\(=x^2+2.5x+5^2\)
\(=\left(x+5\right)^2\)
\(x^2-6x+9\)
\(=x^2-2.3x+3^2\)
\(=\left(x-3\right)^2\)
\(x^2-x+\dfrac{1}{4}\)
\(=x^2-2.\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}\right)^2\)
\(x^2y^4-2xy^2+1\)
\(=\left(xy^2\right)^2-2.1xy^2+1^2\)
\(=\left(xy^2-1\right)^2\)